# Deriviation of WKB approximation

1. Mar 10, 2007

### Repetit

Hey!

In deriving the WKB approximation the wave function is written as

$$\psi \left( x \right) = exp\left[ i S\left( x \right) \right ]$$

Now, in some of the deriviations I've seen, the function S(x) is expanded as a power series in $$\hbar$$ as

$$S(x) = S_0(x) + \hbar S_1(x) + \frac{\hbar}{2} S_2(x) ...$$

I don't really understand this. It's something like $$S_0$$ being the classical result and, the next term being a first order quantum correction and so on. But why do you choose to expand in powers of $$\hbar$$? Can somebody explain to me what this is all about?

René

2. Mar 10, 2007

### Tom Mattson

Staff Emeritus
That particular form for $S(x)$ has the correspondence principle built right into it. If you take the limit as $\hbar \rightarrow 0$, you recover the classical result.

3. May 21, 2008

### jhicks

(This thread appeared on Google and I have the exact same question) I am extremely confused at your statement. $\hbar$ is a constant, right? How on earth can one construct a power series of a function S(x) by expanding it as a function of a constant? What does that even mean?

I have taken a few (more like 1.5 and some self study) classes in QM on the engineering side, but this is over my head. I've currently borrowed a few different QM textbooks and they all say the same opaque thing.

Last edited: May 21, 2008
4. May 21, 2008

### lbrits

You'll be seeing a lot more constants being treated like parameters in physics, so you'll have to get used to it.

Lets parameterize all the possible universes by different values of $$\hbar$$, and solve quantum mechanics in each of them. Then you'd get a family of solutions parameterized by $$\hbar$$. If you choose our universe, corresponding to our $$\hbar$$, then in principle you have the solution to QM in our universe, no?

There's a caveat, of course. You have to assume that the solutions to problems behave smoothly with $$\hbar$$, which is a reasonable assumption, but only comes from experience.

Anyway, if you stick around long enough you'll get to differentiate with respect to orbital angular momentum $$\ell$$ and all sorts of goodness (Feynman-Hellman theorem)

5. May 21, 2008

### jhicks

I think I understand a little better now, but I'll try to explain what is bugging me still. After reading around, I've come to the conclusion a power series with respect to constants is not so far-fetched: For example, any decimal number can be expressed as a power series in 10, or any other number really.

However, the textbook I am primarily using ( Bransden & Joachain Quantum Mechanics: Second Edition ) mentions the power series for S(x) "does not converge, but is an asymptotic series for the function S(x). As a result, the best approximation to S(x) is obtained by keeping a finite number of terms". I've been reading about asymptotic series, but their rationale/use isn't very clear to me still.

6. May 21, 2008

### jhicks

I guess the closest I can come to what an asymptotic series means in words, it's "Adding more terms to the expansion won't make the relative error appreciably smaller". Is this accurate?